3.2623 \(\int \frac {(5-x) (3+2 x)^{5/2}}{(2+5 x+3 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=175 \[ -\frac {2 (139 x+121) (2 x+3)^{3/2}}{9 \left (3 x^2+5 x+2\right )^{3/2}}+\frac {20 (431 x+364) \sqrt {2 x+3}}{9 \sqrt {3 x^2+5 x+2}}+\frac {11300 \sqrt {-3 x^2-5 x-2} F\left (\sin ^{-1}\left (\sqrt {3} \sqrt {x+1}\right )|-\frac {2}{3}\right )}{9 \sqrt {3} \sqrt {3 x^2+5 x+2}}-\frac {8620 \sqrt {-3 x^2-5 x-2} E\left (\sin ^{-1}\left (\sqrt {3} \sqrt {x+1}\right )|-\frac {2}{3}\right )}{9 \sqrt {3} \sqrt {3 x^2+5 x+2}} \]

[Out]

-2/9*(3+2*x)^(3/2)*(121+139*x)/(3*x^2+5*x+2)^(3/2)+20/9*(364+431*x)*(3+2*x)^(1/2)/(3*x^2+5*x+2)^(1/2)-8620/27*
EllipticE(3^(1/2)*(1+x)^(1/2),1/3*I*6^(1/2))*(-3*x^2-5*x-2)^(1/2)*3^(1/2)/(3*x^2+5*x+2)^(1/2)+11300/27*Ellipti
cF(3^(1/2)*(1+x)^(1/2),1/3*I*6^(1/2))*(-3*x^2-5*x-2)^(1/2)*3^(1/2)/(3*x^2+5*x+2)^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {818, 820, 843, 718, 424, 419} \[ -\frac {2 (139 x+121) (2 x+3)^{3/2}}{9 \left (3 x^2+5 x+2\right )^{3/2}}+\frac {20 (431 x+364) \sqrt {2 x+3}}{9 \sqrt {3 x^2+5 x+2}}+\frac {11300 \sqrt {-3 x^2-5 x-2} F\left (\sin ^{-1}\left (\sqrt {3} \sqrt {x+1}\right )|-\frac {2}{3}\right )}{9 \sqrt {3} \sqrt {3 x^2+5 x+2}}-\frac {8620 \sqrt {-3 x^2-5 x-2} E\left (\sin ^{-1}\left (\sqrt {3} \sqrt {x+1}\right )|-\frac {2}{3}\right )}{9 \sqrt {3} \sqrt {3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

Int[((5 - x)*(3 + 2*x)^(5/2))/(2 + 5*x + 3*x^2)^(5/2),x]

[Out]

(-2*(3 + 2*x)^(3/2)*(121 + 139*x))/(9*(2 + 5*x + 3*x^2)^(3/2)) + (20*Sqrt[3 + 2*x]*(364 + 431*x))/(9*Sqrt[2 +
5*x + 3*x^2]) - (8620*Sqrt[-2 - 5*x - 3*x^2]*EllipticE[ArcSin[Sqrt[3]*Sqrt[1 + x]], -2/3])/(9*Sqrt[3]*Sqrt[2 +
5*x + 3*x^2]) + (11300*Sqrt[-2 - 5*x - 3*x^2]*EllipticF[ArcSin[Sqrt[3]*Sqrt[1 + x]], -2/3])/(9*Sqrt[3]*Sqrt[2
+ 5*x + 3*x^2])

Rule 419

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1*EllipticF[ArcSin[Rt[-(d/c),
2]*x], (b*c)/(a*d)])/(Sqrt[a]*Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &
& GtQ[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-(b/a), -(d/c)])

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 718

Int[((d_.) + (e_.)*(x_))^(m_)/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(2*Rt[b^2 - 4*a*c, 2]
*(d + e*x)^m*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))])/(c*Sqrt[a + b*x + c*x^2]*((2*c*(d + e*x))/(2*c*d -
b*e - e*Rt[b^2 - 4*a*c, 2]))^m), Subst[Int[(1 + (2*e*Rt[b^2 - 4*a*c, 2]*x^2)/(2*c*d - b*e - e*Rt[b^2 - 4*a*c,
2]))^m/Sqrt[1 - x^2], x], x, Sqrt[(b + Rt[b^2 - 4*a*c, 2] + 2*c*x)/(2*Rt[b^2 - 4*a*c, 2])]], x] /; FreeQ[{a, b
, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && EqQ[m^2, 1/4]

Rule 818

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1)*(2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g
- c*(b*e*f + b*d*g + 2*a*e*g))*x))/(c*(p + 1)*(b^2 - 4*a*c)), x] - Dist[1/(c*(p + 1)*(b^2 - 4*a*c)), Int[(d +
e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Simp[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a
*e*(e*f*(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*(m + p + 1) + 2*c^2*d*f*(m +
2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2*p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && Ne
Q[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] &&
RationalQ[a, b, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 820

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*(f*b - 2*a*g + (2*c*f - b*g)*x))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/
((p + 1)*(b^2 - 4*a*c)), Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1)*Simp[g*(2*a*e*m + b*d*(2*p + 3)) - f*
(b*e*m + 2*c*d*(2*p + 3)) - e*(2*c*f - b*g)*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &&
NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] && (IntegerQ[m] || IntegerQ[p]
|| IntegersQ[2*m, 2*p])

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {(5-x) (3+2 x)^{5/2}}{\left (2+5 x+3 x^2\right )^{5/2}} \, dx &=-\frac {2 (3+2 x)^{3/2} (121+139 x)}{9 \left (2+5 x+3 x^2\right )^{3/2}}+\frac {2}{9} \int \frac {(-480-145 x) \sqrt {3+2 x}}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx\\ &=-\frac {2 (3+2 x)^{3/2} (121+139 x)}{9 \left (2+5 x+3 x^2\right )^{3/2}}+\frac {20 \sqrt {3+2 x} (364+431 x)}{9 \sqrt {2+5 x+3 x^2}}-\frac {4}{9} \int \frac {1820+2155 x}{\sqrt {3+2 x} \sqrt {2+5 x+3 x^2}} \, dx\\ &=-\frac {2 (3+2 x)^{3/2} (121+139 x)}{9 \left (2+5 x+3 x^2\right )^{3/2}}+\frac {20 \sqrt {3+2 x} (364+431 x)}{9 \sqrt {2+5 x+3 x^2}}-\frac {4310}{9} \int \frac {\sqrt {3+2 x}}{\sqrt {2+5 x+3 x^2}} \, dx+\frac {5650}{9} \int \frac {1}{\sqrt {3+2 x} \sqrt {2+5 x+3 x^2}} \, dx\\ &=-\frac {2 (3+2 x)^{3/2} (121+139 x)}{9 \left (2+5 x+3 x^2\right )^{3/2}}+\frac {20 \sqrt {3+2 x} (364+431 x)}{9 \sqrt {2+5 x+3 x^2}}-\frac {\left (8620 \sqrt {-2-5 x-3 x^2}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1+\frac {2 x^2}{3}}}{\sqrt {1-x^2}} \, dx,x,\frac {\sqrt {6+6 x}}{\sqrt {2}}\right )}{9 \sqrt {3} \sqrt {2+5 x+3 x^2}}+\frac {\left (11300 \sqrt {-2-5 x-3 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+\frac {2 x^2}{3}}} \, dx,x,\frac {\sqrt {6+6 x}}{\sqrt {2}}\right )}{9 \sqrt {3} \sqrt {2+5 x+3 x^2}}\\ &=-\frac {2 (3+2 x)^{3/2} (121+139 x)}{9 \left (2+5 x+3 x^2\right )^{3/2}}+\frac {20 \sqrt {3+2 x} (364+431 x)}{9 \sqrt {2+5 x+3 x^2}}-\frac {8620 \sqrt {-2-5 x-3 x^2} E\left (\sin ^{-1}\left (\sqrt {3} \sqrt {1+x}\right )|-\frac {2}{3}\right )}{9 \sqrt {3} \sqrt {2+5 x+3 x^2}}+\frac {11300 \sqrt {-2-5 x-3 x^2} F\left (\sin ^{-1}\left (\sqrt {3} \sqrt {1+x}\right )|-\frac {2}{3}\right )}{9 \sqrt {3} \sqrt {2+5 x+3 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.34, size = 196, normalized size = 1.12 \[ -\frac {\frac {17240 \left (3 x^2+5 x+2\right )}{\sqrt {2 x+3}}-\frac {6 \sqrt {2 x+3} \left (12930 x^3+32192 x^2+26161 x+6917\right )}{3 x^2+5 x+2}-\frac {1840 (x+1) \sqrt {\frac {3 x+2}{2 x+3}} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {5}{3}}}{\sqrt {2 x+3}}\right )|\frac {3}{5}\right )}{\sqrt {\frac {x+1}{10 x+15}}}+\frac {8620 (x+1) \sqrt {\frac {3 x+2}{2 x+3}} E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {5}{3}}}{\sqrt {2 x+3}}\right )|\frac {3}{5}\right )}{\sqrt {\frac {x+1}{10 x+15}}}}{27 \sqrt {3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((5 - x)*(3 + 2*x)^(5/2))/(2 + 5*x + 3*x^2)^(5/2),x]

[Out]

-1/27*((17240*(2 + 5*x + 3*x^2))/Sqrt[3 + 2*x] - (6*Sqrt[3 + 2*x]*(6917 + 26161*x + 32192*x^2 + 12930*x^3))/(2
+ 5*x + 3*x^2) + (8620*(1 + x)*Sqrt[(2 + 3*x)/(3 + 2*x)]*EllipticE[ArcSin[Sqrt[5/3]/Sqrt[3 + 2*x]], 3/5])/Sqr
t[(1 + x)/(15 + 10*x)] - (1840*(1 + x)*Sqrt[(2 + 3*x)/(3 + 2*x)]*EllipticF[ArcSin[Sqrt[5/3]/Sqrt[3 + 2*x]], 3/
5])/Sqrt[(1 + x)/(15 + 10*x)])/Sqrt[2 + 5*x + 3*x^2]

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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (4 \, x^{3} - 8 \, x^{2} - 51 \, x - 45\right )} \sqrt {3 \, x^{2} + 5 \, x + 2} \sqrt {2 \, x + 3}}{27 \, x^{6} + 135 \, x^{5} + 279 \, x^{4} + 305 \, x^{3} + 186 \, x^{2} + 60 \, x + 8}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)*(3+2*x)^(5/2)/(3*x^2+5*x+2)^(5/2),x, algorithm="fricas")

[Out]

integral(-(4*x^3 - 8*x^2 - 51*x - 45)*sqrt(3*x^2 + 5*x + 2)*sqrt(2*x + 3)/(27*x^6 + 135*x^5 + 279*x^4 + 305*x^
3 + 186*x^2 + 60*x + 8), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {{\left (2 \, x + 3\right )}^{\frac {5}{2}} {\left (x - 5\right )}}{{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)*(3+2*x)^(5/2)/(3*x^2+5*x+2)^(5/2),x, algorithm="giac")

[Out]

integrate(-(2*x + 3)^(5/2)*(x - 5)/(3*x^2 + 5*x + 2)^(5/2), x)

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maple [B]  time = 0.03, size = 308, normalized size = 1.76 \[ \frac {2 \left (77580 x^{4}+309522 x^{3}+1293 \sqrt {15}\, \sqrt {2 x +3}\, \sqrt {-2 x -2}\, \sqrt {-30 x -20}\, x^{2} \EllipticE \left (\frac {\sqrt {30 x +45}}{5}, \frac {\sqrt {15}}{3}\right )+402 \sqrt {15}\, \sqrt {2 x +3}\, \sqrt {-2 x -2}\, \sqrt {-30 x -20}\, x^{2} \EllipticF \left (\frac {\sqrt {30 x +45}}{5}, \frac {\sqrt {15}}{3}\right )+446694 x^{2}+2155 \sqrt {15}\, \sqrt {2 x +3}\, \sqrt {-2 x -2}\, \sqrt {-30 x -20}\, x \EllipticE \left (\frac {\sqrt {30 x +45}}{5}, \frac {\sqrt {15}}{3}\right )+670 \sqrt {15}\, \sqrt {2 x +3}\, \sqrt {-2 x -2}\, \sqrt {-30 x -20}\, x \EllipticF \left (\frac {\sqrt {30 x +45}}{5}, \frac {\sqrt {15}}{3}\right )+276951 x +862 \sqrt {2 x +3}\, \sqrt {15}\, \sqrt {-2 x -2}\, \sqrt {-30 x -20}\, \EllipticE \left (\frac {\sqrt {30 x +45}}{5}, \frac {\sqrt {15}}{3}\right )+268 \sqrt {2 x +3}\, \sqrt {15}\, \sqrt {-2 x -2}\, \sqrt {-30 x -20}\, \EllipticF \left (\frac {\sqrt {30 x +45}}{5}, \frac {\sqrt {15}}{3}\right )+62253\right ) \sqrt {3 x^{2}+5 x +2}}{27 \left (x +1\right )^{2} \left (3 x +2\right )^{2} \sqrt {2 x +3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5-x)*(2*x+3)^(5/2)/(3*x^2+5*x+2)^(5/2),x)

[Out]

2/27*(402*15^(1/2)*EllipticF(1/5*(30*x+45)^(1/2),1/3*15^(1/2))*x^2*(2*x+3)^(1/2)*(-2*x-2)^(1/2)*(-30*x-20)^(1/
2)+1293*15^(1/2)*EllipticE(1/5*(30*x+45)^(1/2),1/3*15^(1/2))*x^2*(2*x+3)^(1/2)*(-2*x-2)^(1/2)*(-30*x-20)^(1/2)
+670*15^(1/2)*EllipticF(1/5*(30*x+45)^(1/2),1/3*15^(1/2))*x*(2*x+3)^(1/2)*(-2*x-2)^(1/2)*(-30*x-20)^(1/2)+2155
*15^(1/2)*EllipticE(1/5*(30*x+45)^(1/2),1/3*15^(1/2))*x*(2*x+3)^(1/2)*(-2*x-2)^(1/2)*(-30*x-20)^(1/2)+268*(2*x
+3)^(1/2)*15^(1/2)*(-2*x-2)^(1/2)*(-30*x-20)^(1/2)*EllipticF(1/5*(30*x+45)^(1/2),1/3*15^(1/2))+862*(2*x+3)^(1/
2)*15^(1/2)*(-2*x-2)^(1/2)*(-30*x-20)^(1/2)*EllipticE(1/5*(30*x+45)^(1/2),1/3*15^(1/2))+77580*x^4+309522*x^3+4
46694*x^2+276951*x+62253)*(3*x^2+5*x+2)^(1/2)/(x+1)^2/(3*x+2)^2/(2*x+3)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {{\left (2 \, x + 3\right )}^{\frac {5}{2}} {\left (x - 5\right )}}{{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)*(3+2*x)^(5/2)/(3*x^2+5*x+2)^(5/2),x, algorithm="maxima")

[Out]

-integrate((2*x + 3)^(5/2)*(x - 5)/(3*x^2 + 5*x + 2)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ -\int \frac {{\left (2\,x+3\right )}^{5/2}\,\left (x-5\right )}{{\left (3\,x^2+5\,x+2\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((2*x + 3)^(5/2)*(x - 5))/(5*x + 3*x^2 + 2)^(5/2),x)

[Out]

-int(((2*x + 3)^(5/2)*(x - 5))/(5*x + 3*x^2 + 2)^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)*(3+2*x)**(5/2)/(3*x**2+5*x+2)**(5/2),x)

[Out]

Timed out

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